15+4x-3x^2=0

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Solution for 15+4x-3x^2=0 equation: 



15+4x-3x^2=0

a = -3; b = 4; c = +15;
Δ = b2-4ac
Δ = 42-4·(-3)·15
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*-3}=\frac{-18}{-6}
=+3
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*-3}=\frac{10}{-6}
=-1+2/3
$

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